I'm looking for a way to solve this problem visually, e.g. by looking at a map and directly marking on it. In my example the majority of vertices (37 out of 45) have odd order. So listing all possible pairs isn't feasible! However, it seems like almost all pairings will obviously not result in the shortest path, e.g. it should only be necessary to look at some amount of adjacent vertices. Any thoughts on such a visual algorithm?
Blimey. 37 odd nodes! No, I can see why you don't want to list them all! I am sorry but I don't know a way of doing that. It may be possible to write something for a computer to solve, but that is beyond my expertise. Sorry.
@@homunculus3646 I am slightly confused as the number 37 doesn't appear in the video at all! There are 4 odd nodes in the example that I have shown. The numbers such as 39 and 25 and so on are representing the total weight of the arcs in a given path and so can be any total at all, depending on the weights of the arcs.
@@mathshelpwithmrorys8555 it was in response to the previous comment from michael smith. He said that his graph had 37 nodes with an odd order which is impossible (by the handshaking lemma)
@@homunculus3646 Ah, yes. A reply to a reply. I couldn't see them when I replied to yours. Sorry. And you are right - I should have picked up on the fact that you can only have an even number of odd order nodes. Thank you for picking me up on my oversight.
hi may I know which apps you used to create this video
because i want to do my assignment presentation for chinese postman problem too
Hi, it is done with PowerPoint. God luck with the assignment.
@@mathshelpwithmrorys8555 thank you
Just do on paper like I did
ua-cam.com/video/Q-eihhuToGo/v-deo.html
I'm looking for a way to solve this problem visually, e.g. by looking at a map and directly marking on it. In my example the majority of vertices (37 out of 45) have odd order. So listing all possible pairs isn't feasible! However, it seems like almost all pairings will obviously not result in the shortest path, e.g. it should only be necessary to look at some amount of adjacent vertices. Any thoughts on such a visual algorithm?
Blimey. 37 odd nodes! No, I can see why you don't want to list them all! I am sorry but I don't know a way of doing that. It may be possible to write something for a computer to solve, but that is beyond my expertise. Sorry.
in a connected graph it is impossible to have an odd number of odd nodes without loops so idk how you have 37
@@homunculus3646 I am slightly confused as the number 37 doesn't appear in the video at all! There are 4 odd nodes in the example that I have shown. The numbers such as 39 and 25 and so on are representing the total weight of the arcs in a given path and so can be any total at all, depending on the weights of the arcs.
@@mathshelpwithmrorys8555 it was in response to the previous comment from michael smith. He said that his graph had 37 nodes with an odd order which is impossible (by the handshaking lemma)
@@homunculus3646 Ah, yes. A reply to a reply. I couldn't see them when I replied to yours. Sorry. And you are right - I should have picked up on the fact that you can only have an even number of odd order nodes. Thank you for picking me up on my oversight.
Thanks
You're welcome.